A. Space Navigation
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard output
You were dreaming that you are traveling to a planet named Planetforces on your personal spaceship. Unfortunately, its piloting system was corrupted and now you need to fix it in order to reach Planetforces.

Space can be represented as the XY plane. You are starting at point (0,0), and Planetforces is located in point (px,py).

The piloting system of your spaceship follows its list of orders which can be represented as a string s. The system reads s from left to right. Suppose you are at point (x,y) and current order is si:

if si=U, you move to (x,y+1);
if si=D, you move to (x,y−1);
if si=R, you move to (x+1,y);
if si=L, you move to (x−1,y).
Since string s could be corrupted, there is a possibility that you won’t reach Planetforces in the end. Fortunately, you can delete some orders from s but you can’t change their positions.

Can you delete several orders (possibly, zero) from s in such a way, that you’ll reach Planetforces after the system processes all orders?

Input
The first line contains a single integer t (1≤t≤1000) — the number of test cases.

Each test case consists of two lines. The first line in each test case contains two integers px and py (−105≤px,py≤105; (px,py)≠(0,0)) — the coordinates of Planetforces (px,py).

The second line contains the string s (1≤|s|≤105: |s| is the length of string s) — the list of orders.

It is guaranteed that the sum of |s| over all test cases does not exceed 105.

Output
For each test case, print “YES” if you can delete several orders (possibly, zero) from s in such a way, that you’ll reach Planetforces. Otherwise, print “NO”. You can print each letter in any case (upper or lower).

Example
inputCopy
6
10 5
RRRRRRRRRRUUUUU
1 1
UDDDRLLL
-3 -5
LDLDLDDDR
1 2
LLLLUU
3 -2
RDULRLLDR
-1 6
RUDURUUUUR
outputCopy
YES
YES
YES
NO
YES
NO
Note
In the first case, you don’t need to modify s, since the given s will bring you to Planetforces.

In the second case, you can delete orders s2, s3, s4, s6, s7 and s8, so s becomes equal to “UR”.

In the third test case, you have to delete order s9, otherwise, you won’t finish in the position of Planetforces.

本题很简单，对于二维坐标系来说，从原点到任意位置，本题处理过程中只是要求该点的可到达性，如果某些步骤如果影响可到达性那么只需要忽略这个步骤即可，那么就是判断四种方向的累计和是不是不小于对应点的坐标的绝对值即可。
需要注意的是，分类讨论的时候需要把不同情况分类清楚即可
本题也可以逆向去做，直接在可到达点根据相应的操作进行相减即可
代码：

#include<bits/stdc++.h>
using namespace std;
#define FAST ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
int main()
{
    
    
    FAST;
    int n;
    int t;
    cin>>t;
    while(t--)
    {
    
    
        bool flag=false;
        int x,y;
        cin>>x>>y;
        string a;
        int b[4];//保存四个方向的累计和
        cin>>a;
        memset(b,0,sizeof(b));
        for(int i=0;i<a.size();i++)
        {
    
    
            if(a[i]=='R')  b[0]++;
            if(a[i]=='L')  b[1]++;
            if(a[i]=='U')  b[2]++;
            if(a[i]=='D')  b[3]++;
        }
        if(x>0&&y>0)  {
    
    //分类讨论
            if(b[0]>=x&&b[2]>=y)
            {
    
    
                flag=true;
            }
        }
        if(x>0&&y<0){
    
    
            y=(-1)*y;
            if(b[0]>=x&&b[3]>=y)
            {
    
    
                flag=true;
            }
        }
        if(x<0&&y>0){
    
    
            x=(-1)*x;
            if(b[1]>=x&&b[2]>=y)
            {
    
    
                flag=true;
            }

        }
        if(x<0&&y<0){
    
    
            x=(-1)*x;
            y=(-1)*y;
            if(b[1]>=x&&b[3]>=y){
    
    
                flag=true;
            }
        } 
        if(x==0&&y>0)  {
    
    
            if(b[2]>=y)
            {
    
    
                flag=true;
            }
        }
        if(x==0&&y<0){
    
    
            y=(-1)*y;
            if(b[3]>=y)
            {
    
    
                flag=true;
            }
        }
        if(x>0&&y==0){
    
    
            if(b[0]>=x)
            {
    
    
                flag=true;
            }

        }
        if(x<0&&y==0){
    
    
            x=(-1)*x;
            if(b[1]>=x){
    
    
                flag=true;
            }
        } 
        if(flag)
        {
    
    
            cout<<"YES"<<endl;
        }
        else
        {
    
    
            cout<<"NO"<<endl;
        }
        flag=false;
    }
   //system("pause");
}